任务:打印从1到1000的数字,不使用任何循环或条件语句。不要只写printf()或cout语句1000次。

用C或c++怎么做呢?


当前回答

使用系统命令:

system("/usr/bin/seq 1000");

其他回答

#include <boost/mpl/range_c.hpp>
#include <boost/mpl/for_each.hpp>
#include <boost/lambda/lambda.hpp>
#include <iostream>

int main()
{
  boost::mpl::for_each<boost::mpl::range_c<unsigned, 1, 1001> >(std::cout << boost::lambda::_1 << '\n');
  return(0);
}
#include <stdio.h>
int main() { printf("numbers from 1 to 1000"); return 0; }

这就像另一个以“愤怒”结尾的英语单词谜语,对吧?

这是我的两个解。第一个是c#,第二个是C语言:

C#:

const int limit = 1000;

Action<int>[] actions = new Action<int>[2];
actions[0] = (n) => { Console.WriteLine(n); };
actions[1] = (n) => { Console.WriteLine(n);  actions[Math.Sign(limit - n-1)](n + 1); };

actions[1](0);

C:

#define sign(x) (( x >> 31 ) | ( (unsigned int)( -x ) >> 31 ))

void (*actions[3])(int);

void Action0(int n)
{
    printf("%d", n);
}

void Action1(int n)
{
    int index;
    printf("%d\n", n);
    index = sign(998-n)+1;
    actions[index](++n);
}

void main()
{
    actions[0] = &Action0;
    actions[1] = 0; //Not used
    actions[2] = &Action1;

    actions[2](0);
}
    static void Main(string[] args)
    {
        print(1000);
        System.Console.ReadKey();
    }

    static bool print(int val)
    {
        try
        {
            print( ((val/val)*val) - 1);
            System.Console.WriteLine(val.ToString());
        }
        catch (Exception ex)
        {
            return false;
        }
        return true;
    }
#include <iostream>
#include <vector>

using namespace std;
#define N 10    //10 or 1000, doesn't matter

class A{
public:
    A(){
        //cout << "A(): " << m_ << endl;    //uncomment to show the difference between gcc and Microsoft C++ compiler
    }
    A(const A&){
        ++m_;
        cout << m_ << endl;     
    }
private:
    static int m_;  //global counter
};

int A::m_(0);  //initialization

int main(int argc, char* argv[])
{
    //Creates a vector with N elements. Printing is from the copy constructor, 
    //which is called exactly N times.
    vector<A> v(N);  
    return 0;   
}

实现注意: 使用gcc:默认构造函数创建一个“master”元素。 然后元素被复制构造函数复制N次。 在微软c++编译器中:所有元素都是由默认构造函数创建的 然后被复制构造函数复制。